Question

The Goodyear blimps, which frequently fly over sporting events, hold approximately \(175,000 \mathrm{ft}^{3}\) of helium. If the gas is at \(23^{\circ} \mathrm{C}\) and \(1.0 \mathrm{~atm}\), what mass of helium is in a blimp?

Short answer

The mass of helium in the Goodyear blimp is approximately 880,705.12 g, calculated using the Ideal Gas Law, given the volume of \(175,000 ft^3\), temperature of \(23^\circ\mathrm{C}\), and pressure of \(1.0 \mathrm{~atm}\).

Step-by-step solution

Convert given temperature to Kelvin

To use the Ideal Gas Law, the temperature should be in Kelvin. The given temperature is 23°C. To convert Celsius to Kelvin, add 273.15: \( T = 23^\circ\mathrm{C} + 273.15 = 296.15\mathrm{K} \)

Apply Ideal Gas Law to calculate moles

Now, we will use the Ideal Gas Law formula: PV = nRT Given, pressure P = 1.0 atm, volume V = 175,000 ft³, temperature T = 296.15 K, and ideal gas constant R = 0.0821 L atm / K mol. Note that we need to convert the volume from ft³ to L before using the formula. There are approximately 28.3168 L in 1 ft³: \( V = 175,000\mathrm{ft^3} * \frac{28.3168\mathrm{L}}{1\mathrm{ft^3}} = 4,955,440\mathrm{L} \) Now, plug in the values into the Ideal Gas Law formula and solve for n: \( (1.0\mathrm{atm}) (4,955,440\mathrm{L}) = n (0.0821\frac{\mathrm{L\,atm}}{\mathrm{K\,mol}})(296.15\mathrm{K}) \) Divide both sides by 0.0821 L atm / K mol and 296.15 K: \( n = \frac{(1.0\mathrm{atm})(4,955,440\mathrm{L})}{(0.0821\frac{\mathrm{L\,atm}}{\mathrm{K\,mol}})(296.15\mathrm{K})} \) Calculate the number of moles, n: \( n \approx 219916.22\, \mathrm{mol} \)

Calculate the mass of helium

We now have the number of moles of helium. To find the mass, we will use the molar mass of helium, which is 4.0026 g/mol: Mass = moles × molar mass \( \mathrm{Mass} = (219916.22\, \mathrm{mol}) (4.0026\, \frac{\mathrm{g}}{\mathrm{mol}}) \) Calculate the mass of helium: \( \mathrm{Mass} \approx 880705.12\, \mathrm{g} \) The mass of helium in the blimp is approximately 880,705.12 g.